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(A) The magnetic force on a current-carrying wire segment is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Given $\vec{B} = B\hat{j} = 0.02\hat{j}\,T

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(A) The magnetic force on a current-carrying wire segment is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Given $\vec{B} = B\hat{j} = 0.02\hat{j}\,T$.For segment $ab$: The vector $\vec{L}_{ab}$ is along the negative $y$-axis,so $\vec{L}_{ab} = -0.4\hat{j}\,m$. The force $\vec{F}_{ab} = I(\vec{L}_{ab} 	imes \vec{B}) = I(-0.4\hat{j} 	imes 0.02\hat{j}) = 0$,since the cross product of parallel vectors is zero.For segment $bc$: The vector $\vec{L}_{bc}$ is along the $x$-axis,so $\vec{L}_{bc} = 0.4\hat{i}\,m$. The force $\vec{F}_{bc} = I(\vec{L}_{bc} 	imes \vec{B}) = I(0.4\hat{i} 	imes 0.02\hat{j}) = I(0.008\hat{k})\,N$.The magnitude of force on $ab$ is $F_{ab} = 0$,and the magnitude of force on $bc$ is $F_{bc} = 5 	imes 0.008 = 0.04\,N$.The ratio of the magnetic force on segment $ab$ to the magnetic force on segment $bc$ is $0 / 0.04 = 0$.

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